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Anton [14]
3 years ago
8

There are 1,295 students attending a small university.There are 714 women enrolled.What percent of the students are women?

Mathematics
1 answer:
BARSIC [14]3 years ago
3 0
Total students = 1295

Number of women = 714

Percentage of women = 714/1295 x 100 = 55.14%

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Answer: 55.14% of the students are women.
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we know that
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2) we know that

a/sinA=b/sin B=c/sinC

and

∡K=α

∡M=β

ME=b

then

b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))

KE=b*sin(β)/sin(α)

A=(1/2)*(ME)*(KE)*sin(180-(α+β))

sin(180-(α+β))=sin(α+β)

A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

KE/sin(β)=KM/sin(180-(α+β))

KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)

the answers part 2) are

side KE=b*sin(β)/sin(α)
side KM=(KE/sin(β))*sin(α+β)
Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

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